Solve $x^3-y^3-z^3=3^t cdot 2xyz$

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Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$



We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.



I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.










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  • This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
    – Right
    Aug 30 at 6:29










  • @Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
    – Oleg567
    Aug 30 at 6:34










  • @Oleg567 That is right, did not see it immediately.
    – Right
    Aug 30 at 6:36














up vote
6
down vote

favorite
2












Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$



We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.



I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.










share|cite|improve this question























  • This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
    – Right
    Aug 30 at 6:29










  • @Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
    – Oleg567
    Aug 30 at 6:34










  • @Oleg567 That is right, did not see it immediately.
    – Right
    Aug 30 at 6:36












up vote
6
down vote

favorite
2









up vote
6
down vote

favorite
2






2





Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$



We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.



I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.










share|cite|improve this question















Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$



We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.



I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.







number-theory diophantine-equations






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share|cite|improve this question













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edited Aug 30 at 5:55

























asked Aug 30 at 2:45









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8,46932269











  • This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
    – Right
    Aug 30 at 6:29










  • @Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
    – Oleg567
    Aug 30 at 6:34










  • @Oleg567 That is right, did not see it immediately.
    – Right
    Aug 30 at 6:36
















  • This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
    – Right
    Aug 30 at 6:29










  • @Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
    – Oleg567
    Aug 30 at 6:34










  • @Oleg567 That is right, did not see it immediately.
    – Right
    Aug 30 at 6:36















This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
– Right
Aug 30 at 6:29




This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
– Right
Aug 30 at 6:29












@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
– Oleg567
Aug 30 at 6:34




@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
– Oleg567
Aug 30 at 6:34












@Oleg567 That is right, did not see it immediately.
– Right
Aug 30 at 6:36




@Oleg567 That is right, did not see it immediately.
– Right
Aug 30 at 6:36










1 Answer
1






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oldest

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up vote
3
down vote



accepted










Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.



Some solutions for small $t$'s.

beginarrayc
hline
t& u & v & w \ hline
2 & 35917476 & 6829645 & 10182731\ hline
3 & 1137565 & 647349 & 30196 \ hline
4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
endarray
It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.



for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.



For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)



In particular at $t=6$ the original curve
$$
frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
$$
is birational to
$$
y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
$$
via
$$
beginalign
(u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
endalign
$$




We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.



Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.




$t=2$ example



For $t=2$, we want integer solutions to
$$
f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
$$
We may freely multiply a rational $tneq 0$ to get
$$
tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
$$
Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).



Let
$$
beginalign
t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
endalign
$$
then it may be verified that this transform the original equation to
$$
g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
$$
which is a projective elliptic curve.
(In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)



Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
$$
E: y^2 = x^3-36288x+2285712
$$
This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.




Sample solutions.



The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
$$
(-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
$$
Converting to $(u,v,w)$, we get
$$
(frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
$$
Normalizing, the first three solutions are
$$
beginalign
(95,-42,-13),\
(-35917476, -6829645, -10182731)\
(54059017558123943, -5902844861231317, -35013190193908290)
endalign
$$



Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
$$
(35917476, 6829645, 10182731)
$$






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.



    Some solutions for small $t$'s.

    beginarrayc
    hline
    t& u & v & w \ hline
    2 & 35917476 & 6829645 & 10182731\ hline
    3 & 1137565 & 647349 & 30196 \ hline
    4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
    endarray
    It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.



    for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.



    For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)



    In particular at $t=6$ the original curve
    $$
    frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
    $$
    is birational to
    $$
    y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
    $$
    via
    $$
    beginalign
    (u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
    endalign
    $$




    We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.



    Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.




    $t=2$ example



    For $t=2$, we want integer solutions to
    $$
    f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
    $$
    We may freely multiply a rational $tneq 0$ to get
    $$
    tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
    $$
    Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).



    Let
    $$
    beginalign
    t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
    u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
    endalign
    $$
    then it may be verified that this transform the original equation to
    $$
    g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
    $$
    which is a projective elliptic curve.
    (In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)



    Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
    $$
    E: y^2 = x^3-36288x+2285712
    $$
    This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.




    Sample solutions.



    The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
    $$
    (-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
    $$
    Converting to $(u,v,w)$, we get
    $$
    (frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
    $$
    Normalizing, the first three solutions are
    $$
    beginalign
    (95,-42,-13),\
    (-35917476, -6829645, -10182731)\
    (54059017558123943, -5902844861231317, -35013190193908290)
    endalign
    $$



    Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
    $$
    (35917476, 6829645, 10182731)
    $$






    share|cite|improve this answer


























      up vote
      3
      down vote



      accepted










      Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.



      Some solutions for small $t$'s.

      beginarrayc
      hline
      t& u & v & w \ hline
      2 & 35917476 & 6829645 & 10182731\ hline
      3 & 1137565 & 647349 & 30196 \ hline
      4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
      endarray
      It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.



      for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.



      For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)



      In particular at $t=6$ the original curve
      $$
      frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
      $$
      is birational to
      $$
      y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
      $$
      via
      $$
      beginalign
      (u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
      endalign
      $$




      We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.



      Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.




      $t=2$ example



      For $t=2$, we want integer solutions to
      $$
      f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
      $$
      We may freely multiply a rational $tneq 0$ to get
      $$
      tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
      $$
      Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).



      Let
      $$
      beginalign
      t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
      u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
      endalign
      $$
      then it may be verified that this transform the original equation to
      $$
      g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
      $$
      which is a projective elliptic curve.
      (In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)



      Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
      $$
      E: y^2 = x^3-36288x+2285712
      $$
      This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.




      Sample solutions.



      The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
      $$
      (-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
      $$
      Converting to $(u,v,w)$, we get
      $$
      (frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
      $$
      Normalizing, the first three solutions are
      $$
      beginalign
      (95,-42,-13),\
      (-35917476, -6829645, -10182731)\
      (54059017558123943, -5902844861231317, -35013190193908290)
      endalign
      $$



      Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
      $$
      (35917476, 6829645, 10182731)
      $$






      share|cite|improve this answer
























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.



        Some solutions for small $t$'s.

        beginarrayc
        hline
        t& u & v & w \ hline
        2 & 35917476 & 6829645 & 10182731\ hline
        3 & 1137565 & 647349 & 30196 \ hline
        4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
        endarray
        It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.



        for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.



        For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)



        In particular at $t=6$ the original curve
        $$
        frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
        $$
        is birational to
        $$
        y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
        $$
        via
        $$
        beginalign
        (u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
        endalign
        $$




        We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.



        Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.




        $t=2$ example



        For $t=2$, we want integer solutions to
        $$
        f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
        $$
        We may freely multiply a rational $tneq 0$ to get
        $$
        tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
        $$
        Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).



        Let
        $$
        beginalign
        t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
        u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
        endalign
        $$
        then it may be verified that this transform the original equation to
        $$
        g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
        $$
        which is a projective elliptic curve.
        (In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)



        Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
        $$
        E: y^2 = x^3-36288x+2285712
        $$
        This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.




        Sample solutions.



        The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
        $$
        (-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
        $$
        Converting to $(u,v,w)$, we get
        $$
        (frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
        $$
        Normalizing, the first three solutions are
        $$
        beginalign
        (95,-42,-13),\
        (-35917476, -6829645, -10182731)\
        (54059017558123943, -5902844861231317, -35013190193908290)
        endalign
        $$



        Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
        $$
        (35917476, 6829645, 10182731)
        $$






        share|cite|improve this answer














        Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.



        Some solutions for small $t$'s.

        beginarrayc
        hline
        t& u & v & w \ hline
        2 & 35917476 & 6829645 & 10182731\ hline
        3 & 1137565 & 647349 & 30196 \ hline
        4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
        endarray
        It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.



        for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.



        For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)



        In particular at $t=6$ the original curve
        $$
        frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
        $$
        is birational to
        $$
        y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
        $$
        via
        $$
        beginalign
        (u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
        endalign
        $$




        We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.



        Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.




        $t=2$ example



        For $t=2$, we want integer solutions to
        $$
        f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
        $$
        We may freely multiply a rational $tneq 0$ to get
        $$
        tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
        $$
        Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).



        Let
        $$
        beginalign
        t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
        u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
        endalign
        $$
        then it may be verified that this transform the original equation to
        $$
        g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
        $$
        which is a projective elliptic curve.
        (In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)



        Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
        $$
        E: y^2 = x^3-36288x+2285712
        $$
        This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.




        Sample solutions.



        The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
        $$
        (-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
        $$
        Converting to $(u,v,w)$, we get
        $$
        (frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
        $$
        Normalizing, the first three solutions are
        $$
        beginalign
        (95,-42,-13),\
        (-35917476, -6829645, -10182731)\
        (54059017558123943, -5902844861231317, -35013190193908290)
        endalign
        $$



        Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
        $$
        (35917476, 6829645, 10182731)
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 31 at 7:53

























        answered Aug 30 at 7:53









        Yong Hao Ng

        2,7791219




        2,7791219



























             

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